Упростите и вычислите: ∛(9+√80)+∛(9-√80)=

[latex]9+\sqrt{80}=9+4\sqrt5=\frac18\cdot(72+32\sqrt5)=\frac18\cdot(27+27\sqrt5+45+5\sqrt5)=\\=\frac18\cdot(3^3+3\cdot3^2\cdot\sqrt5+3\cdot3\cdot(\sqrt5)^2+(\sqrt5)^3)=\frac18\cdot(3+\sqrt5)^3\\\\9-\sqrt{80}=9-4\sqrt5=\frac18\cdot(72-32\sqrt5)=\frac18\cdot(27-27\sqrt5+45-5\sqrt5)=\\=\frac18\cdot(3^3-3\cdot3^2\cdot\sqrt5+3\cdot3\cdot(\sqrt5)^2-(\sqrt5)^3)=\frac18\cdot(3-\sqrt5)^3\\\\\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=\sqrt[3]{\frac{(3+\sqrt5)^3}8}+\sqrt[3]{\frac{(3-\sqrt5)^3}8}=[/latex] [latex]=\frac{3+\sqrt5}2+\frac{3-\sqrt5}2=\frac{3+\sqrt5+3-\sqrt5}2=\frac62=3[/latex]